To solve this problem we will use the formula for compound interest:
[tex]P_N=P_0\cdot(1+\frac{r}{k})^N\text{.}[/tex]Where:
• P_N is the balance in the account after N years,
,• P_0 is the starting balance of the account (also called an initial deposit, or principal),
,• r is the annual interest rate in decimal form,
,• N in years,
,• k is the number of compounding periods in one year.
In this problem, we have:
• P_0 = $328,120,,
,• interest P_N - P_0 = $23,515.27 → ,P_N = $351,635.27,,
,• N = ,200 days = ,200/365,
,• k = 1.
From the formula above, we have:
[tex]\begin{gathered} (\frac{P_N}{P_0})^{\frac{1}{N}}=1+r \\ r=(\frac{P_N}{P_0})^{\frac{1}{N}}-1. \end{gathered}[/tex]Replacing the data of the problem, we get:
[tex]r=(\frac{351,635.27}{328,120})^{\frac{365}{200}}-1\cong0.1346\cong13.5％.[/tex]Answer
The annual interest is 13.5%.
